Please help Determine the final concentration of each reactant in the mixed solu

Question

Please help Determine the final concentration of each reactant in the mixed solutions in each procedure Record these values in Table 1. Plot the data from the data tables at the end of the lab as x-v data points either in Excel (or an equivalent spreadsheet program) Your graphs) should present data similar to that contained in the graph shown below. Using Excel find the slope of the line using the linear regression function (a separate graph with just the points on the line will probably need to be made to do this.) Record the slope of the line for each procedure as the Initial Rate in Table 1. Using the correct rate law from the first page calculate the rale constant for each procedure and record it in Table 1. Convert the procedure temperatures into units Kelvin and record those values in Table 1. Plot the converted values In k vs. 1/T. Determine the activation energy E. in kJ/mol. (use the correct value for R - 0.008314 kJ/mol K). Determine the value for A. Determine the activat

Explanation / Answer

problem 11) solution -

Arrhenius equation,

k=Aexp(-Ea/RT)

or, ln k=ln A-Ea/RT

where k=rate constant,A=frequency factor ,T=temperature ,Ea=activation energy ,R=universal gas constant

for two different temperatures ,T1 and T2 ,let the rate constants for a rxn be k1,k2 respectively

ln k1=ln A-Ea/RT1...............(1)

ln k2=ln A-Ea/RT2................(2)

eqn (1)-(2)gives

ln(k2/k1)=Ea/R(1/T1-1/T2)

or, using data from the table k1=0.000108

k2=0.001599

k3=0.00250

T1=280.35K

T2=310.85K

T3=294.55K

ln 0.001599/0.000108=Ea/8.314J/K.mol(1/280.35K - 1/310.85K).................(1)

solving eqn (1),

2.695=Ea/8.314J/K.mol(0.00357K-1 - 0.00322K-1)..

or, 2.695=Ea(0.0000425(J/mol)^-1

Ea=63411.765 J/mol

putting this value in eqn (1),

ln 0.000108=ln A-[63411.765/(8.314J/K.mol)280.35K]

or,-9.133=lnA-27.206

lnA=18.072

A=7.06*10^7

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